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--- lecture_10_slides [2016/02/24 16:51] – [Example] rupert
+++ lecture_10_slides [2017/02/21 10:04] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
-==== Last time ====
-  * $A\vec{x}=\vec{b}$ is equivalent to a system of linear equations
+===== Matrix equations =====
-  * More generally, we might want to solve \[AX=B\] where
-    * $A$, $B$ are fixed matrices
+  * A linear equation can be written using [[row-column multiplication]].
-    * $X$ is an unknown matrix we want to find.
+  * e.g. $ \newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]} 2x-3y+z=8$ is same as $ \m{2&-3&1}\m{x\\y\\z}=8$
-  * Can figure out the size of $X$ from the size of $A$ and $B$
+  * or $ a\vec x=8$ where $a=\m{2&-3&1}$ and $\vec x=\m{x\\y\\z}$.
-  * To solve: find the matrices $X$. How?
-  * Can't "divide by $A$" (can't divide by a matrix...)
+==== ====
+  * We can write a whole [[system of linear equations]] in a similar way, as a matrix equation using [[matrix multiplication]].
+  * e.g. the linear system $\begin{align*} 2x-3y+z&=8\\ y-z&=4\\x+y+z&=0\end{align*}$
+  * is same as $\m{2&-3&1\\0&1&-1\\1&1&1}\m{x\\y\\z}=\m{8\\4\\0}$
+  * or $ A\vec x=\vec b$ where $A=\m{2&-3&1\\0&1&-1\\1&1&1}$, $\vec x=\m{x\\y\\z}$ and $\vec b=\m{8\\4\\0}$.
+==== ====
+In a similar way, any linear system \begin{align*} a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_m&=b_1\\ a_{21}x_1+a_{22}x_2+\dots+a_{2m}x_m&=b_2\\ \hphantom{a_{11}}\vdots \hphantom{x_1+a_{22}}\vdots\hphantom{x_2+\dots+{}a_{nn}} \vdots\ & \hphantom{{}={}\!} \vdots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nm}x_m&=b_n \end{align*}
+can be written in the form
+\[ A\vec x=\vec b\]
+where $A$ is the $n\times m $ matrix, called the **coefficient matrix** of the linear system, whose $(i,j)$ entry is $a_{ij}$ (the number in front of $x_j$ in the $i$th equation of the system).
+==== Solutions of matrix equations ====
+  * More generally, might want to solve a matrix equation like \[AX=B\] where $A$, $X$ and $B$ are matrices of any size, with $A$ and $B$ fixed matrices and $X$ a matrix of unknown variables.
+  * If $A$ is $n\times m$, we need $B$ to be $n\times k$ for some $k$, and then $X$ must be $m\times k$
+    * so we know the size of any solution $X$.
+  *  But which $m\times k$ matrices $X$ are solutions?
+==== Example ====
+If $A=\m{1&0\\0&0}$ and $B=0_{2\times 3}$, then any solution $X$ to $AX=B$ must be $2\times 3$.
+  * One solution is $X=0_{2\times 3}$
+    *  because then we have $AX=A0_{2\times 3}=0_{2\times 3}$.
+  * This is not the only solution!
+  * For example, $X=\m{0&0&0\\1&2&3}$ is another solution
+    * because then we have $AX=\m{1&0\\0&0}\m{0&0&0\\1&2&3}=\m{0&0&0\\0&0&0}=0_{2\times 3}.$
+  * So a matrix equation can have more than one solution.
+==== Example ====
+  * Let $A=\m{2&4\\0&1}$
+  * and $B=\m{3&4\\5&6}$
+  * Solve $AX=B$ for $X$
+  * $X$ must be $2\times 2$
+  * $X=\m{x_{11}&x_{12}\\x_{21}&x_{22}}$
+  * Do some algebra to solve for $X$
+  * ...
+  * Is there a quicker way?
 ==== Example ====
@@ Line 150: / Line 192: @@
 ==== Example ====
-  * Let $A=\mat{1&2\\-3&-6}$. We can now explain why $A$ isn't invertible.
+  * Let $A=\mat{1&2\\-3&-6}$. $A$ isn't invertible... why?
-  * Notice that one column of $A$ is $2$ times the other... exploit this.
+  * One column of $A$ is $2$ times the other... exploit this.
   * Let $K=\mat{-2\\1}$
-  * $K$ is non-zero but $AK=\mat{0\\0}=0_{2\times 1}$
+  * $K$ is non-zero but $AK=\mat{1&2\\-3&-6}\mat{-2\\1}=\mat{0\\0}=0_{2\times 1}$
   * So $A$ is not invertible, by the Corollary.
-  * Next time: an easier way to determine when a matrix is invertible: **determinants**
+  * Next time: a more systematic way to determine when a matrix is invertible: **determinants**